Problem: What is the extraneous solution to these equations? $\dfrac{x^2}{x + 6} = \dfrac{36}{x + 6}$
Solution: Multiply both sides by $x + 6$ $ \dfrac{x^2}{x + 6} (x + 6) = \dfrac{36}{x + 6} (x + 6)$ $ x^2 = 36$ Subtract $36$ from both sides: $ x^2 - (36) = 36 - (36)$ $ x^2 - 36 = 0$ Factor the expression: $ (x + 6)(x - 6) = 0$ Therefore $x = -6$ or $x = 6$ At $x = -6$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -6$, it is an extraneous solution.